Find the derivatives of the following at any point of their domains :
`(x+1)/x`

Answer

To find the derivative of the function \( f(x) = \frac{x+1}{x} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

### Step-by-step Solution:

1. **Identify \( u \) and \( v \)**:
- Here, \( u = x + 1 \) and \( v = x \).

2. **Find the derivatives of \( u \) and \( v \)**:
- \( \frac{du}{dx} = \frac{d}{dx}(x + 1) = 1 \)
- \( \frac{dv}{dx} = \frac{d}{dx}(x) = 1 \)

3. **Apply the quotient rule**:
- Using the quotient rule:
\[ \frac{d}{dx}\left(\frac{x+1}{x}\right) = \frac{x \cdot \frac{du}{dx} - (x + 1) \cdot \frac{dv}{dx}}{x^2} \]
- Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ = \frac{x \cdot 1 - (x + 1) \cdot 1}{x^2} \]

4. **Simplify the expression**:
- This simplifies to:
\[ = \frac{x - (x + 1)}{x^2} \]
- Further simplifying gives:
\[ = \frac{x - x - 1}{x^2} = \frac{-1}{x^2} \]

5. **State the final result**:
- Therefore, the derivative of the function \( f(x) = \frac{x+1}{x} \) is:
\[ f'(x) = -\frac{1}{x^2} \]